Question 129135: I am having some problems with this problem...
Biting an unpopped kernel of popcorn hurts! As an experiemnt, a self-confessed connoisseur of cheap popcorn carefully counted 773 kernels and put them in a popper. After popping, the unpopped kernels were counted. There are 86.
(a) Construct a 90 percent confidence interval for the proportion of all kernels that would not pop.
(b) Check the normality assumption.
(c) Try the Very Quick Rule. Does it work well here?Why or why not?
(d) Why might this sample not be typical?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! As an experiment, a self-confessed connoisseur of cheap popcorn carefully counted 773 kernels and put them in a popper. After popping, the unpopped kernels were counted. There are 86.
---------
p-hat = 86/773=0.1113
std = sqrt[0.1113*0.8887/773]= 0.113
------------
(a) Construct a 90 percent confidence interval for the proportion of all kernels that would not pop.
E = 1.645*0.0113 = 0.0186
C.I.: (p-hat-E , p-hat+E) or (0.0927 , 0.1299)
-------------------
(b) Check the normality assumption.
Check pn>5 and qn>5
0.1113*773 = 86-plus ; q*p-hat is even larger
-------------------------
(c) Try the Very Quick Rule. Does it work well here?Why or why not?
I am not familiar with that Rule.
-----------------
(d) Why might this sample not be typical?
Since the C.I. is a statement about the whole population of popcorn
kernnels, the sample here might not be representative.
---------------
Cheers,
Stan H.
|
|
|
